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Question: What is the derivative of \({{\cot }^{-1}}\left( x \right)\) ?...

What is the derivative of cot⁡−1(x){{\cot }^{-1}}\left( x \right) ?

Explanation

Solution

To find the derivative of cot⁡−1(x){{\cot }^{-1}}\left( x \right) , we have to equate y to cot⁡−1(x){{\cot }^{-1}}\left( x \right) . From this, we will get cot⁡y=x\cot y=x . We have to differentiate this equation with respect to x. Then, we have to use the trigonometric identities to solve further. Finally, we will have to use above substitutions.

Complete step by step solution:
We have to find the derivative of cot⁡−1(x){{\cot }^{-1}}\left( x \right) . Let us first equate y to cot⁡−1(x){{\cot }^{-1}}\left( x \right) .
⇒y=cot⁡−1(x)...(i)\Rightarrow y={{\cot }^{-1}}\left( x \right)...\left( i \right)
We can write the above form as
⇒cot⁡y=x...(ii)\Rightarrow \cot y=x...\left( ii \right)
Let us differentiate the above equation with respect to x. We know that derivative of cot⁡x\cot x is −csc⁡2x-{{\csc }^{2}}x and ddxxn=nxn−1\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} . Therefore, differentiation of the above equation gives
⇒−csc⁡2ydydx=1\Rightarrow -{{\csc }^{2}}y\dfrac{dy}{dx}=1
Let us take −csc⁡2y-{{\csc }^{2}}y to the RHS.
⇒dydx=−1csc⁡2y...(iii)\Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{{{\csc }^{2}}y}...\left( iii \right)
We know that
csc⁡2x−cot⁡2x=1{{\csc }^{2}}x-{{\cot }^{2}}x=1
We can rearrange the terms of this equation so that we can get the identity for csc⁡2x{{\csc }^{2}}x .
⇒csc⁡2x=1+cot⁡2x\Rightarrow {{\csc }^{2}}x=1+{{\cot }^{2}}x
We have to substitute the above formula in equation (iii).
⇒dydx=−11+cot⁡2y\Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{1+{{\cot }^{2}}y}
Let us substitute for cot⁡y\cot y from equation (ii) in the above equation.
⇒dydx=−11+x2\Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{1+{{x}^{2}}}
We can now substitute for y from equation (i) in the above equation.
⇒ddxcot⁡−1x=−11+x2\Rightarrow \dfrac{d}{dx}{{\cot }^{-1}}x=-\dfrac{1}{1+{{x}^{2}}}

Hence, the derivative of cot⁡−1(x){{\cot }^{-1}}\left( x \right) is −11+x2-\dfrac{1}{1+{{x}^{2}}}

Note: Students must never miss out to substitute for cot⁡y\cot y in the final step of differentiation. The final answer must be in terms of x. They must know the trigonometric identities and differentiation of basic functions. Students should never forget to write dydx\dfrac{dy}{dx} when differentiating cot⁡y\cot y with respect to x. They may forget to put the negative sign in the derivative of cot⁡y\cot y .